A ball is dropped from the top of the Empire State Building. Find the velocity of the ball at time t.

Question

A ball is dropped from the top of the Empire State Building to the ground below. The height, y, of the ball above the ground (in metres) is given as funktion of time, t, (in seconds) by y(t) = 380- 5t^2.

Find the velocity of the ball at time t. What is the given sign of the velocity? Why is this to be expected? Show that the acceleration of the ball is a constant. What are the value and sign of this constant? When does the ball hit the ground, and how fast is it going at that time? Give your answers in metres per second and kilometer per hour.

Ich verstehe nicht, warum man bei der Geschwindigkeit ein negatives Vorzeichen erwartet.

Comments

Weil wir uns in Negativer y-Achsenrichtung bewegen.

Die y-Koordinaten sprich die Höhe nimmt hier ja ab. Wird also weniger.

Answer 1

A ball is dropped from the top of the Empire State Building to the ground below. The height, y, of the ball above the ground (in metres) is given as funktion of time, t, (in seconds) by y(t) = 380- 5t² . Find the velocity of the ball at time t. 

v(t) = y'(t) = -10 t

What is the given sign of the velocity? Why is this to be expected? 

Das Vorzeichen ist negativ. Das erwarten wir auch weil wir uns entgegen der y-Achse bewegen.

Show that the acceleration of the ball is a constant. What are the value and sign of this constant? 

a(t) = v'(t) = - 10

When does the ball hit the ground, and how fast is it going at that time? Give your answers in metres per second and kilometer per hour.

y(t) = 0 --> t = 8.718

v(8.718) = - 87.18 m/s = - 314 km/h